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PilotLPTM [1.2K]
3 years ago
6

While camping, you make a snack over a campfire. You roast a marshmallow over the campfire until it is browned and put it betwee

n some chocolate and graham crackers. When you take a bite, you notice that the chocolate and marshmallow have melted slightly. Which statement correctly identifies a part of this process as a physical or a chemical change?
Chemistry
2 answers:
pshichka [43]3 years ago
6 0

Answer:

It is <u>melting</u> which is a physical change.

Explanation:

.

SSSSS [86.1K]3 years ago
6 0

Answer:

It is melting process, hope it help full

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A helium balloon with a volume of 550mL is cooled from 305 to 265K. The pressure on the gas is reduced from 0.45 atm to 0.25 atm
Aleksandr [31]

860 mL.

<h3>Explanation</h3>

Separate this process into two steps:

  1. Cool the balloon from 305 K to 265 K.
  2. Reduce the pressure on the balloon from 0.45 atm to 0.25 atm.

What would be the volume of the balloon after each step?

After Cooling the balloon at constant pressure:

By Charles's Law, the volume of a gas is directly related to its temperature in degrees Kelvins.

In other words,

\dfrac{V_2}{V_1} = \dfrac{T_2}{T_1},

where

  • V_1 and V_2 are volumes of the same gas.
  • T_1 and T_2 are the temperatures (in degrees Kelvins) of that gas.

Rearranging,

V_2 = V_1 \cdot \dfrac{T_2}{T_1}\\\phantom{V_2} = 550 \times \dfrac{265}{305}\\\phantom{V_2} = 478 \; \text{mL}.

The balloon ended up with a lower temperature. As a result, its volume drops: V_2 < V_1.

After reducing the pressure on the balloon at constant temperature:

By Boyle's Law, the volume of a gas is inversely proportional to the pressure on this gas.

In other words,

\dfrac{V_2}{V_1} = \dfrac{P_1}{P_2},

where

  • V_1 and V_2 are volumes of the same gas.
  • P_1 and P_2 are the pressures on this gas.

Rearranging,

V_2 = V_1 \cdot \dfrac{P_1}{P_2}\\\phantom{V_2} = 478 \times \dfrac{0.45}{0.25}\\\phantom{V_2} = 860 \;\text{mL}.

There's now less pressure on the balloon. As a result, the balloon will gain in volume: V_2 > V_1.

The final volume of the balloon will be 860 \; \text{mL}.

7 0
3 years ago
I NEED HELP PLEASE, THANKS!
Crazy boy [7]

Answer:

Here's what I get.

Explanation:

1. Brønsted-Lowry theory

An acid is a substance that can donate a proton to another substance.

A  base is a substance that can accept a proton from another substance.

2. pH of ammonia

The chemical equation is

\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \text{OH}$^{-}$

For simplicity, let's re-write this as

\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$

(a) Set up an ICE table.

                     B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹:     0.335             0        0

C/mol·L⁻¹:       -x                +x       +x

E/mol·L⁻¹:  0.335 + x          x         x

\rm K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.335 - x} = 1.8 \times 10^{-5}

Check for negligibility:

\dfrac{0.335}{1.8 \times 10^{-5}} = 28 000 > 400\\\\x \ll 0.335

(b) Solve for [OH⁻]

\dfrac{x^{2}}{0.335} = 1.8 \times 10^{-5}\\\\x^{2} = 0.335 \times 1.8 \times 10^{-5}\\x^{2} = 6.03 \times 10^{-6}\\x = \sqrt{6.03 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.46 \times 10^{-3}} \textbf{ mol/L}

(c) Calculate the pOH

\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.46 \times 10^{-3}) = 2.61

(d) Calculate the pH

pH = 14.00 - pOH = 14.00 - 2.61 = 11.39

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Gravitational potential energy, mechanical energy, and kinetic energy.
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